Post a New Question


posted by .

An object of mass 4.40 kg is projected into the air at a 55 degre angle. it hits the ground 3.8 s later. what is its change in momentum while it is in the air? ignore air resistance.

  • Physics -

    Impulse = M*g*(3.8 s), downward.
    The launch angle and velocity are not needed for the answer, but they do determine when the object hits the ground. .

  • Physics -

    that doesn't work as the correct answer :/

  • Physics -

    The launch velocity is 22.73 m/s, with a vertical component of 18.6 m/s. It his the top of the trajectory at 18.6/g = 1.90 s and hits the ground at 3.80 s.

    If you use my suggested method, you get
    momentum change = 4.40 kg*9.8 m/s^2*3.8s
    = 163.9 kg m/s

    Initial vertical momentum =
    4.4*sin55*22.73 = 81/93 kg*m/s

    That gets double on the way dwn, for a net momentum change of 163.9 kg m/s

    That answer should work. Your course's "correct" answer may be wrong.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question